Arithmetic of Complex Numbers

Intuition

The previous lesson enlarged the number system. It adjoined one new symbol $i$ , gave it the single property $i^2 = -1$ , and called anything of the form $a + bi$ , with $a$ and $b$ real, a complex number. That was the easy half. A new kind of number is worthless until you can calculate with it, because calculating is what numbers are for. This lesson delivers the arithmetic, and the pleasant surprise is how little of it is new. You add and subtract complex numbers by combining like terms, exactly as with $3 + 5x$ . You multiply them by expanding, exactly as with binomials, and then spend the one new fact $i^2 = -1$ wherever it appears. Even division, the operation that looks hardest, is a move you already own: the same conjugate trick that once cleared $\sqrt{2}$ out of a denominator, pointed at a new target.

Learning goals

Derive every operation from $i^2 = -1$ plus ordinary algebra
Add and subtract part by part
Expand a product, then exchange $i^2$ for $-1$
Use $z\overline{z} = a^2 + b^2$ , real and never negative
Show conjugation passing through sums and products
Divide by rationalizing with the denominator's conjugate

One new symbol, no new laws

When the previous lesson called $a + bi$ a number, that word carried a promise: the new numbers obey the same laws of algebra as the old ones. Addition is still commutative and associative, multiplication still distributes over addition, and subtraction and division still mean what they always meant, namely adding an opposite and multiplying by a reciprocal. The complex numbers add exactly one fact to that rulebook:

i^2 = -1.

Everything in this lesson is forced by that single stipulation. The working strategy throughout is to treat $a + bi$ as a two-term expression in the symbol $i$ and run the ordinary algebra you have used for years. Then exchange $i^2$ for $-1$ the moment it appears. The question worth asking in advance is whether the answers stay inside the system. If you add, subtract, multiply, or divide two complex numbers, is the result again of the form $(\text{real}) + (\text{real})\,i$ ? Watch for the answer as each operation is built, because it will be yes all four times. That is what makes the complex numbers a self-contained place to do mathematics rather than a notational accident.

Before computing anything, pin down what it means for two complex answers to agree. Your work will produce expressions like $23 + 2i$ , and you need to know that no other pair of real numbers can name that same value.

Equal complex numbers have equal parts#

Suppose $a + bi = c + di$ , where all four of $a$ , $b$ , $c$ , $d$ are real. Write $r = a - c$ and $s = b - d$ , both real. Subtracting the right side from the left turns the supposed equality into $r + si = 0$ , that is, $r = -si$ .

Square both sides. The left side becomes $r^2$ , and the right side becomes $(-si)^2 = s^2 i^2 = -s^2$ , so

r^2 = -s^2, \qquad \text{that is,} \qquad r^2 + s^2 = 0.

Now $r$ and $s$ are real numbers, so $r^2 \ge 0$ and $s^2 \ge 0$ . If either of them were nonzero, its square would be strictly positive and the sum $r^2 + s^2$ would be strictly positive, not zero. So both squares are zero, which forces $r = 0$ and $s = 0$ , meaning $a = c$ and $b = d$ .

The converse direction is immediate: if $a = c$ and $b = d$ , the two expressions are the same expression. So two complex numbers are equal exactly when their real parts match and their imaginary parts match.

This little theorem is what makes complex answers readable. When a computation ends at $23 + 2i$ , the real part $23$ and the imaginary part $2$ are not one description among many; they are the only description. It also hands you a technique: an equation between complex numbers is really two equations between real numbers, one for the real parts and one for the imaginary parts. Several problems in the practice sets are solved by exactly that splitting.

Addition and subtraction combine like terms

Add two complex numbers by regrouping, which the commutative and associative laws permit, and then factoring $i$ out of the two imaginary terms, which the distributive law permits:

(a + bi) + (c + di) = (a + c) + (bi + di) = (a + c) + (b + d)\,i.

Real parts add to real parts, and imaginary parts add to imaginary parts, precisely the way $x$ -terms and constant terms keep to themselves when you simplify $(3 + 5x) + (2 + 4x)$ . Subtraction is the same computation after distributing the minus sign through both terms of the second number:

(a + bi) - (c + di) = (a - c) + (b - d)\,i.

Notice what the right sides look like. Since $a + c$ and $b + d$ are real numbers, each result is again $(\text{real}) + (\text{real})\,i$ : adding and subtracting complex numbers can never produce anything but a complex number. Two operations down, two to go.

Worked example 1 Adding and subtracting

Compute $(3 - 5i) + (7 + 2i)$ , then $(4 + 3i) - (6 - 8i)$ .

For the sum, collect the real parts and the imaginary parts separately:

(3 - 5i) + (7 + 2i) = (3 + 7) + (-5 + 2)\,i = 10 - 3i.

For the difference, first distribute the minus sign through the second number, turning $-(6 - 8i)$ into $-6 + 8i$ , and only then collect:

(4 + 3i) - (6 - 8i) = 4 + 3i - 6 + 8i = (4 - 6) + (3 + 8)\,i = -2 + 11i.

The single most common slip in this entire lesson happens in that second computation: subtracting only the $6$ and forgetting that the minus sign also flips the $-8i$ . Distribute first, then combine.

Check your understanding

Compute $(5 - 3i) - (2 + 4i)$ .

3 - 7i

3 + i

7 - 7i

3 + 7i

Multiplication is expansion plus one substitution

Multiply two complex numbers the way you multiply any two binomials: distribute every term of the first factor across every term of the second. Nothing about that step is new. The new step comes after, when an $i^2$ is sitting in the result and must be exchanged for $-1$ :

(a + bi)(c + di) = ac + ad\,i + bc\,i + bd\,i^2 = (ac - bd) + (ad + bc)\,i.

Read the middle expression carefully. The four products are exactly the four you would write for $(a + bx)(c + dx)$ . The only difference is the last one, where $bd\,i^2 = -bd$ abandons the imaginary column and joins the real part with its sign flipped. That is the whole novelty of complex multiplication. Do not memorize the boxed-looking formula $(ac - bd) + (ad + bc)\,i$ ; it is faster and safer to expand and substitute each time, letting the formula rebuild itself. And once more the result is $(\text{real}) + (\text{real})\,i$ , so multiplication also stays inside the system.

An area model for the product of 2 + 3i and 4 - 5i. Each cell holds one of the four products from the expansion. Three cells are routine, but the bottom-right cell contains i squared, and the rule i squared = -1 folds it back into the real column. That returned +15 is why the real part of the answer is 23 rather than 8.

Worked example 2 Two products in full

Compute $(2 + 3i)(4 - 5i)$ , then $(3 - 2i)^2$ .

Expand the first product term by term, exactly as the area model above lays it out:

(2 + 3i)(4 - 5i) = 8 - 10i + 12i - 15i^2.

Now spend the rule. Since $i^2 = -1$ , the last term is $-15i^2 = +15$ , a real number:

8 - 10i + 12i + 15 = (8 + 15) + (-10 + 12)\,i = 23 + 2i.

For the square, use the binomial pattern $(x - y)^2 = x^2 - 2xy + y^2$ with $x = 3$ and $y = 2i$ , or simply expand:

(3 - 2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i.

In both computations the expansion was ordinary algebra; the only complex-number moment was trading each $i^2$ for $-1$ before collecting.

Check your understanding

Compute $(3 + i)(2 - 5i)$ .

6 - 13i

11 - 13i

11 - 17i

1 - 13i

The conjugate makes products real

One special product deserves its own name, because it powers everything in the rest of the lesson. The conjugate of $z = a + bi$ is the number

\overline{z} = a - bi,

the same number with the sign of its imaginary part flipped. So $\overline{4 + 7i} = 4 - 7i$ , the conjugate of $-3 - 5i$ is $-3 + 5i$ , and $\overline{6i} = -6i$ . Flipping the sign twice restores the original, so $\overline{\overline{z}} = z$ : conjugation is its own undo. A real number is its own conjugate, since $r = r + 0i$ and flipping the sign of $0$ changes nothing, and the matching-parts theorem gives the converse too: $\overline{z} = z$ demands $-b = b$ , so $b = 0$ . A complex number equals its own conjugate exactly when it is real.

Pair a number with its conjugate and both basic operations produce real answers. The sum is $(a + bi) + (a - bi) = 2a$ , twice the real part, with the imaginary parts cancelling. The product is the important one.

The product of conjugates is real and never negative#

Expand the product of $z = a + bi$ and $\overline{z} = a - bi$ like any other product:

(a + bi)(a - bi) = a^2 - ab\,i + ab\,i - b^2 i^2 = a^2 + b^2.

The two middle terms are opposites and cancel, and the last term flips sign because $i^2 = -1$ . Compare this with the difference-of-squares pattern $(x + y)(x - y) = x^2 - y^2$ that you know from factoring: the pattern fires here too, giving $a^2 - (bi)^2 = a^2 - b^2 i^2$ . That last $i^2$ turns the familiar difference of squares into a sum of squares. So

z\,\overline{z} = a^2 + b^2,

a real number built from two squares of real numbers. Each square is at least $0$ , so $z\,\overline{z} \ge 0$ always. Better still, the sum-of-squares argument from the equality proof applies word for word: $a^2 + b^2 = 0$ forces $a = 0$ and $b = 0$ , and conversely $z = 0$ gives $0^2 + 0^2 = 0$ . So $z\,\overline{z}$ is strictly positive for every complex number except $z = 0$ itself.

Keep the shape of that result in mind: multiplying any complex number by its conjugate erases $i$ entirely and leaves the positive real number $a^2 + b^2$ . The next lesson will draw $a + bi$ as a point in a plane and read $a^2 + b^2$ off the picture geometrically. For this lesson, what matters is that $a^2 + b^2$ is real and, for $z \ne 0$ , positive.

Conjugation has one more property worth proving now, because a famous fact about quadratics will fall out of it at the end of the lesson. Conjugation does not care whether you conjugate before or after doing arithmetic.

Conjugation respects sums and products#

Let $z = a + bi$ and $w = c + di$ be any two complex numbers, with $a, b, c, d$ real.

For sums, compute both orders. Conjugating after adding gives $\overline{z + w} = \overline{(a + c) + (b + d)\,i} = (a + c) - (b + d)\,i$ . Adding the conjugates gives $\overline{z} + \overline{w} = (a - bi) + (c - di) = (a + c) - (b + d)\,i$ . The two agree, so $\overline{z + w} = \overline{z} + \overline{w}$ .

For products, the multiplication formula from earlier gives $zw = (ac - bd) + (ad + bc)\,i$ , so conjugating after multiplying yields

\overline{zw} = (ac - bd) - (ad + bc)\,i.

Multiplying the conjugates first instead,

\overline{z}\,\overline{w} = (a - bi)(c - di) = ac - ad\,i - bc\,i + bd\,i^2 = (ac - bd) - (ad + bc)\,i,

the same number. So $\overline{zw} = \overline{z}\,\overline{w}$ as well. In words: conjugate first or compute first, the answer is identical, for addition and for multiplication. The answer is therefore identical for anything built out of those two operations, such as subtraction, squaring, and higher powers.

Worked example 3 Conjugate products at work

Compute $(3 + 4i)(3 - 4i)$ and $(1 + 2i)(1 - 2i)$ without a full expansion.

Each is a number times its own conjugate, so the theorem answers immediately with $a^2 + b^2$ :

(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 25, \qquad (1 + 2i)(1 - 2i) = 1^2 + 2^2 = 5.

If you distrust the shortcut, expand the first one: $9 - 12i + 12i - 16i^2 = 9 + 16 = 25$ . The cross terms cancel and the $i^2$ flips the last sign, exactly as the proof said they must. Notice both answers are positive real numbers; an $i$ can never survive this pairing.

Division is rationalizing in disguise

You have solved this problem before, in a different costume. When a denominator held a radical, you multiplied by a well-chosen form of $1$ to clear it:

\frac{1}{3 + \sqrt{2}} = \frac{1}{3 + \sqrt{2}} \cdot \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}.

The conjugate $3 - \sqrt{2}$ was chosen because the difference-of-squares pattern makes the new denominator $3^2 - (\sqrt{2})^2 = 7$ , a rational number. Division of complex numbers is the same trick with the same name. To divide by $c + di$ , multiply numerator and denominator by its conjugate $c - di$ , and the proof above guarantees the new denominator is the real number $c^2 + d^2$ :

\frac{a + bi}{c + di} = \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di} = \frac{(a + bi)(c - di)}{c^2 + d^2}.

Multiplying by $\frac{c - di}{c - di}$ is multiplying by $1$ , so the value never changes; only its costume does. After this move, the numerator is one ordinary complex multiplication, and dividing a complex number by the positive real $c^2 + d^2$ just divides each part, since $\frac{p + qi}{m} = \frac{p}{m} + \frac{q}{m}\,i$ .

Pause on when this is allowed. The move needs $c^2 + d^2 \ne 0$ , and the conjugate proof showed $c^2 + d^2 = 0$ happens only for $c + di = 0$ . So the only forbidden division is division by zero, the same single exception the real numbers always had. Every nonzero complex number can divide, and in particular every nonzero $z = c + di$ has a reciprocal,

\frac{1}{c + di} = \frac{c - di}{c^2 + d^2},

again of the form $(\text{real}) + (\text{real})\,i$ . That settles the closure question asked at the start: all four operations take complex numbers to complex numbers. The system that began as one symbol and one rule supports the full arithmetic that the rational and real numbers enjoy, with nothing missing and nothing extra needed.

Worked example 4 A quotient, start to finish

Compute $\dfrac{3 + 2i}{1 - 4i}$ .

The denominator is $1 - 4i$ , so multiply numerator and denominator by its conjugate $1 + 4i$ :

\frac{3 + 2i}{1 - 4i} \cdot \frac{1 + 4i}{1 + 4i} = \frac{(3 + 2i)(1 + 4i)}{(1 - 4i)(1 + 4i)}.

The denominator is a conjugate product, so it is $1^2 + 4^2 = 17$ with no work. Expand the numerator and spend $i^2 = -1$ :

(3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8i^2 = 3 + 14i - 8 = -5 + 14i.

Now divide each part by $17$ :

\frac{3 + 2i}{1 - 4i} = \frac{-5 + 14i}{17} = -\frac{5}{17} + \frac{14}{17}\,i.

The answer is not pretty, but it is exact, and it has the required shape of a real part plus a real multiple of $i$ .

Worked example 5 Division undoes multiplication

Find the complex number $z$ satisfying $(2 - i)\,z = 4 + 3i$ .

Division exists precisely to undo multiplication, so divide both sides by $2 - i$ , which is legal because $2 - i \ne 0$ :

z = \frac{4 + 3i}{2 - i} = \frac{(4 + 3i)(2 + i)}{(2 - i)(2 + i)} = \frac{8 + 4i + 6i + 3i^2}{2^2 + 1^2} = \frac{5 + 10i}{5} = 1 + 2i.

Check it, because a check costs one multiplication:

(2 - i)(1 + 2i) = 2 + 4i - i - 2i^2 = 2 + 3i + 2 = 4 + 3i. \checkmark

Linear equations over the complex numbers solve exactly the way they did over the reals, because the algebraic laws are the same and the reciprocal always exists.

Check your understanding

Compute $\dfrac{2 + i}{3 - i}$ .

\dfrac{1}{2} - \dfrac{1}{2}i

\dfrac{5}{8} + \dfrac{5}{8}i

\dfrac{1}{2} + \dfrac{1}{2}i

5 + 5i

The quadratics that had no solutions

Here is the payoff the chapter has been building toward. In the last chapter, the discriminant sorted every quadratic, and the verdict for $\Delta < 0$ was blunt: no real solutions. Take $z^2 - 4z + 5 = 0$ , whose discriminant is $\Delta = (-4)^2 - 4(1)(5) = -4$ . The quadratic formula asks for a number whose square is $-4$ , and the reals have none; but the complex numbers do, because $(2i)^2 = 4i^2 = -4$ . Feeding $2i$ into the formula in place of the missing square root suggests the candidates

z = \frac{4 \pm 2i}{2} = 2 \pm i.

The formula was derived for real arithmetic, so treat this as a suggestion rather than a conclusion. Then let the arithmetic of this lesson give the honest verdict by direct substitution.

Worked example 6 Checking a complex root

Verify that $z = 2 + i$ satisfies $z^2 - 4z + 5 = 0$ .

Compute the square first, expanding and spending $i^2 = -1$ :

z^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i.

Then assemble the whole left side, distributing $-4$ across $2 + i$ :

z^2 - 4z + 5 = (3 + 4i) - (8 + 4i) + 5 = (3 - 8 + 5) + (4 - 4)\,i = 0.

Both the real part and the imaginary part collapse to $0$ , so $2 + i$ really is a solution, certified by nothing but addition and multiplication.

What about the other candidate, $2 - i$ ? It is the conjugate of the root just checked, and you do not need to repeat the computation to trust it. Conjugate both sides of the true statement $z^2 - 4z + 5 = 0$ . The right side is the real number $0$ , its own conjugate. The left side is built from $z$ by multiplication and addition with the real numbers $-4$ and $5$ . The proof above showed conjugation respects both operations while leaving real numbers fixed, so the conjugate of that left side is $\overline{z}^2 - 4\overline{z} + 5$ . The conjugated statement therefore reads $\overline{z}^2 - 4\overline{z} + 5 = 0$ : whenever a complex number solves this equation, its conjugate automatically solves it too. Notice also that the pair fits the sum and product of roots from the last chapter perfectly: $(2 + i) + (2 - i) = 4$ and $(2 + i)(2 - i) = 2^2 + 1^2 = 5$ , matching $z^2 - 4z + 5$ term for term. The final lesson of this chapter turns these observations into the full story of quadratics with negative discriminants.

Common mistakes

Leaving an $i^2$ unspent. The expansion $(2 + 3i)(4 - 5i) = 8 - 10i + 12i - 15i^2$ is not finished: the last term must become $+15$ and join the real part. Treating $i^2$ as if it were $1$ , or dropping it, is the most common multiplication error, and it changes $23 + 2i$ into a wrong answer like $-7 + 2i$ .
Subtracting only half of the second number. In $(4 + 3i) - (6 - 8i)$ , the minus sign hits both terms: $4 + 3i - 6 + 8i$ . Forgetting to flip the $-8i$ gives $-2 - 5i$ instead of the correct $-2 + 11i$ .
Calling $(a + bi)(a - bi)$ a difference of squares and stopping at $a^2 - b^2$ . The pattern does fire, but it produces $a^2 - b^2 i^2$ , and the $i^2$ flips the sign: the product of conjugates is the sum $a^2 + b^2$ . That plus sign is the entire reason the conjugate clears denominators.
Multiplying only the denominator by the conjugate. To compute $\frac{a + bi}{c + di}$ you multiply by $\frac{c - di}{c - di}$ , a form of $1$ . Multiplying the bottom alone (or using the numerator’s conjugate) changes the value of the fraction, not just its form.
Multiplying square roots of negatives as if they were positive. The rule $\sqrt{x}\,\sqrt{y} = \sqrt{xy}$ is proved only for non-negative $x$ and $y$ . Since $(3i)^2 = 9i^2 = -9$ , the square root of $-9$ is $3i$ , and likewise the square root of $-4$ is $2i$ ; their product is $(3i)(2i) = 6i^2 = -6$ , not $\sqrt{36} = 6$ . Convert to $i$ -form first, then multiply.
Flipping the wrong sign when conjugating. The conjugate changes only the imaginary part: the conjugate of $-3 - 5i$ is $-3 + 5i$ , not $3 + 5i$ . The real part never moves.
Stopping short of the form $a + bi$ . An answer like $\frac{5 + 10i}{5}$ or $8 - 10i + 12i - 15i^2$ is unfinished work. Reduce every power of $i$ , combine like terms, and split any real denominator across both parts before calling a computation done.

Key takeaways

Complex numbers obey the ordinary laws of algebra plus the single rule $i^2 = -1$ ; every operation in this lesson is forced by that stipulation.
Equality is part-by-part: $a + bi = c + di$ exactly when $a = c$ and $b = d$ , so one complex equation is two real equations.
Add and subtract like terms: $(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)\,i$ .
Multiply by expanding, then exchange every $i^2$ for $-1$ : $(a + bi)(c + di) = (ac - bd) + (ad + bc)\,i$ . Re-derive this by expansion rather than memorizing it.
The conjugate of $z = a + bi$ is $\overline{z} = a - bi$ . Then $z + \overline{z} = 2a$ and $z\,\overline{z} = a^2 + b^2$ , which is real, is never negative, and equals $0$ only for $z = 0$ .
Conjugation respects arithmetic: $\overline{z + w} = \overline{z} + \overline{w}$ and $\overline{zw} = \overline{z}\,\overline{w}$ , and it fixes exactly the real numbers.
Divide by rationalizing: multiply numerator and denominator by the conjugate of the denominator, so the denominator becomes the positive real $c^2 + d^2$ ; in particular $\frac{1}{c + di} = \frac{c - di}{c^2 + d^2}$ .
All four operations return answers of the form $(\text{real}) + (\text{real})\,i$ , and only division by $0$ is forbidden, so the complex numbers are closed under the same arithmetic the reals support.
For quadratics with real coefficients, conjugating a true equation shows complex solutions arrive in conjugate pairs, a fact this chapter’s final lesson develops in full.

Arithmetic of Complex Numbers

Learning goals

One new symbol, no new laws

Addition and subtraction combine like terms

Multiplication is expansion plus one substitution

The conjugate makes products real

Division is rationalizing in disguise

The quadratics that had no solutions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)